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3 years ago

Which of the following cycles tracks the movement of water in the hydrosphere?

Chemistry

1 answer:

ratelena [41]3 years ago

8 0

Answer:

The hydrologic cycle

Explanation:

The water cyle is also known as hydrologic cycle describes about movement of water in the hydrosphere.

The nitrogen cycle describes about the movement of nitrogen in atmosphere.

The rock cycle describes about formation and decomposition of rocks.

the hydrogen cycle describes about movement of hydrogen in the atmosphere.

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What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?

Maru [420]

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

Step 1 : Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M

This means there is 0.05 mol per 100mL

Step 2 : calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

3 0

3 years ago

Read 2 more answers

What is the direction of the polarity of the indicated bond in h3c−oh marked by δ+ and δ−?

Sedbober [7]

Answer:- The direction of the polarity of the indicated bond is from carbon to oxygen.

Explanations:- There are two types of covalent compounds, polar and non polar. If the bond is between two same atoms for example, H-H, Cl-Cl etc then the bond is non polar. If the bond is between two different atoms then the bond would be polar. The direction of the polarity is from loss electron negative atom to more electron negative atom.

Oxygen is more electron negative than carbon. So, being more electron negative, the bonding electrons are more towards oxygen and it cases partial negative charge on oxygen and partial positive charge on carbon. The direction of the polarity is from less electron negative carbon to more electron negative oxygen.

It is shown in the diagram below:

6 0

3 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

3 years ago

Chalcopyrite is an ore with the composition cufes2. what is the percentage of iron in a 39.6 g sample of this ore? answer in uni

iren2701 [21]

Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g

5 0

3 years ago

Which reaction below represents a balanced, double replacement chemical reaction?

Mariulka [41]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

I think the reaction that represents a balanced, double replacement chemical reaction is B which is Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO

4 0

4 years ago